Question

Dissociation constant of the reaction, $\left[Ag\right(N{H}_3{)}_2{]}^{+}\rightleftharpoons A{g}^{+}+2N{H}_3$ is $6\times {10}^{-14}.$ Calculate the $E^o$ value of the reaction as $y\times {10}^{-3}eV$, then y is
$\left[Ag\right(N{H}_3{)}_2{]}^{+}+e^{-}\to Ag+2N{H}_3$
[Given :$A{g}^{+}+e^{-}\to Ag;E^0=0.799V$]

Answer 1

$[Ag{(NH}_3{)}_2{]}^{+}\rightleftharpoons {Ag}^{+}+2N{H}_3$
$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}log\frac{\left[A{g}^{+}\right][N{H}_3{]}^2}{\left[Ag\right(N{H}_3{)}_2{]}^{+}}$
$E_{cell}$ = $E_{cell}^{\circ }=\frac{0.059}{1}logK=\frac{0.059}{1}log6\times {10}^{-14}=-0.780$
$E_{\left(Ag\right(N{H}_3{)}_2)/Ag}^{\circ }=-0.780+0.799=0.019V=19\times {10}^{-3}$